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olga55 [171]
2 years ago
13

Which of the following clusters of orbitals could form a shape similar to that shown here (Figure 3) in the valence shell of an

isolated atom or one about to enter into bonding with other atoms?
A) five sp³d
B) three sp² and one p orbital

Chemistry
1 answer:
pav-90 [236]2 years ago
5 0
"three sp²" is the one cluster of orbitals among the following choices given in the question that <span>could form a shape similar to that shown here (Figure 3) in the valence shell of an isolated atom or one about to enter into bonding with other atoms. The correct option among all the options given in the question is the second option.</span>
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How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?
Allushta [10]
I dont know but do you know da wae brudda?


3 0
2 years ago
Naturally occurring iodine has an atomic mass of 126.9045 amu. A 12.3849-g sample of iodine is accidentally contaminated with 1.
Oliga [24]

Answer:

127.0665 amu

Explanation:

Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.

We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g

The percentage abundances is as follows:

Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%

Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%

Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.

That is :

[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu

6 0
2 years ago
If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?
EleoNora [17]
<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.
3 0
2 years ago
Read 2 more answers
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
1 year ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
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